# Construction of Components of Second Bimedial

## Theorem

In the words of Euclid:

To find medial straight lines commensurable in square only which contain a medial rectangle.

## Proof

Let $\rho$, $\rho \sqrt k$ and $\rho \sqrt \lambda$ be rational straight lines which are commensurable in square only.

Take the mean proportional $\rho \sqrt  k$ of $\rho$ and $\rho \sqrt k$, which is medial.

Let $x$ be such that:

$\rho \sqrt k : \rho \sqrt \lambda = \rho \sqrt  k : x$

which gives:

$x = \dfrac {\rho \sqrt \lambda} {\sqrt  k}$

We have that:

$\rho \sqrt k \frown \!\! - \rho \sqrt \lambda$

where $\frown \!\! -$ denotes commensurability in square only.

Thus:

$\rho \sqrt  k \frown \!\! - x$

From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial.

Then:

 $\ds \rho \sqrt \lambda : x$ $=$ $\ds \rho \sqrt k : \rho \sqrt  k$ $\ds$ $=$ $\ds \rho \sqrt  k : \rho$ $\ds \leadsto \ \$ $\ds x \cdot \rho \sqrt  k$ $=$ $\ds \rho^2 \sqrt \lambda$ which is a medial area

$\blacksquare$

## Historical Note

This proof is Proposition $28$ of Book $\text{X}$ of Euclid's The Elements.